How are R^2 and R^4 isomorphic as VS's over Q?

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In summary: The tensor product is a way to extend the field of scalars of a vector space. A real vector space of dimension n (n can be infinite) becomes a complex vector space of dimension n when tensored with C. It becomes a real vector space of twice the dimension over R.However, over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.
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joeblow
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I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R2 and the second is isomorphic to R4 over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)
 
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joeblow said:
I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R2 and the second is isomorphic to R4 over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)

Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.

The tensor product is a way to extend the field of scalars of a vector space. A real vector space of dimension n (n can be infinite) becomes a complex vector space of dimension n when tensored with C. It becomes a real vector space of twice the dimension over R.

If one views R as a 1 dimensional vector space over itself with single basis vector v, the all elements are of the form rV for real numbers,r. Over C the basis is also v and the resulting vector space is 1 dimensional over C. But over R the basis is V and iV and so is 2 dimensional over R.
 
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  • #3
lavinia said:
Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.
Indeed; we discussed this here.
 

FAQ: How are R^2 and R^4 isomorphic as VS's over Q?

How are R2 and R4 isomorphic as VS's over Q?

The vector spaces R2 and R4 are isomorphic over Q because they have the same dimension. This means that there exists a linear transformation from R2 to R4 that is both one-to-one and onto, preserving the vector space structure and operations.

What is the significance of being isomorphic over Q?

Being isomorphic over Q means that two vector spaces have the same algebraic structure, and therefore can be studied and analyzed in a similar way. This allows for the use of techniques and theorems from one vector space to be applied to the other.

Can R2 and R4 be isomorphic over a field other than Q?

Yes, R2 and R4 can be isomorphic over other fields as well. In fact, they are isomorphic over any field with characteristic 0, such as the real numbers, rational numbers, or complex numbers.

How can we prove that R2 and R4 are isomorphic over Q?

One way to prove that R2 and R4 are isomorphic over Q is by constructing an explicit linear transformation between the two vector spaces. This can be done by defining a basis for each vector space and showing that the transformation preserves the operations of addition and scalar multiplication.

Are there any other vector spaces that are isomorphic to R2 and R4 over Q?

Yes, there are infinitely many vector spaces that are isomorphic to R2 and R4 over Q. In general, any vector space with dimension 2 or 4 over Q will be isomorphic to R2 and R4, respectively. This is because isomorphic vector spaces have the same dimension.

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